Section 7.4 The Lax Milgram Lemma
(The presentation is adapted from L.C. Evans' book \cite{Evans98})\\ The Lax Milgram Lemma is not really a fixed-point theorem, but it deals with solutions to nonlinear equations, in particular quadratic operator equations. It is a generalization of Riesz representation theorem.Theorem 7.4.1.
Let \(H\) be a Hilbert space and \(B:H\times H\to \RR\) a continuous, bilinear mapping. We assume that here are two constants \(\alpha, \beta>0\) such that
\begin{align*}
|B(u,v) | \amp\leq\amp \alpha \|u \|\; \|v\| \qquad \;\;\;\mbox{for all }\quad\quad u,v\in H,\\
B(u,u) \amp\geq\amp \beta \|u\|^2 \qquad\qquad \mbox{for all }\quad \quad u\in H.
\end{align*}
The second condition is known as the coercivity condition. Then for each \(f\in H^*\) there exists a solution \(u\in H\) of
\begin{equation*}
B(u,v) = f(v), \qquad \mbox{for all }\quad \quad v\in H.
\end{equation*}
Proof.
- For each fixed \(u\in H\text{,}\) the mapping \(v\mapsto B(u,v)\) is linear and bounded, hence in \(H^*\text{.}\) By the Riesz representation theorem Theorem 5.5.1, there exists a unique representative \(w\in H\) such that\begin{equation*} (w,v) = B(u,v), \qquad \mbox{for all }\quad \quad v\in H. \end{equation*}We write \(Au=w\text{,}\) defining a map \(A:H\to H\) such that\begin{equation*} B(u,v) = (Au, v). \end{equation*}To prove the Lax-Milgram results, we need to show that \(A\) is invertible.
- We show first that \(A:H\to H\) is linear and bounded. By direct computation we get:\begin{equation*} (A(a_1 u_1+ a_2 u_2), v) = B(a_1 u_1+a_2 u_2, v) = a_1 B(u_1, v) + a_2 B(u_2, v) = (a_1 Au_1+ a_2 A u_2, v). \end{equation*}Furthermore\begin{equation*} \|Au\|^2 = (Au, Au) = B(u, Au)\leq \alpha \|u \|\; \|Au\|. \end{equation*}Hence\begin{equation*} \|Au\| \leq \alpha \|u\|\qquad\mbox{and}\qquad \|A\|\leq \alpha. \end{equation*}
- From the coercivity condition we get\begin{equation*} \beta \|u\|^2 \leq B(u,u) = (Au,u) \leq \|Au\|\;\|u\|. \end{equation*}Hence\begin{equation*} \beta\|u\|\leq \|Au\|. \end{equation*}This implies that the mapping \(A\) is injective and the range \(R(A)\) is closed.
- We now show that \(R(A)=H\text{.}\) If not, then since the range is closed, we find a nontrivial orthogonal element \(z\in R(A)^\perp\text{.}\) For this \(z\) we have\begin{equation*} \beta \|z\|^2 \leq B(z,z) = (Az, z) = 0, \end{equation*}which is a contradiction. Hence \(R(A)=H\) and we can invert \(A\text{.}\)
- Then for each \(f\in H^*\text{,}\) we find \(w\in H\) and \(u\in H\) such that\begin{equation*} f(v) = (w,v)=(Au, v) = B(u,v), \end{equation*}which was to be shown.
- \(u\text{.}\)\(u\)\(\tilde u\text{.}\)\(B(u-\tilde u, v) =0\)\(v\in H\text{.}\)\begin{equation*} \beta \|u-\tilde u\|^2 \leq B(u-\tilde u, u-\tilde u) = 0 \end{equation*}\(u=\tilde u\text{.}\)
Example 7.4.2.
\begin{equation*}
L u + \mu u = f
\end{equation*}
where \(L\) is a general second order differential operator. Using weak formulation with test functions \(v\) we write
\begin{equation*}
\underbrace{\int Lu\; v\; dx +\mu \int uvdx}_{=B(u,v)} = \underbrace{\int f v dx}_{=(f,v)},
\end{equation*}
and the problem becomes: given \(f\in X^*\text{,}\) find \(u\in X\) to solve \(B(u,v)=f(v)\text{,}\) i.e. a Lax-Milgram situation. The details of the estimates in Theorem Theorem 7.4.1 need to be worked out, depending on the second order differential operator and possible boundary conditions.